证明|z1+z2|2+|z1-z2|2=2(|z1|2+|z2|2),并说明其几何意义.
[单选题]若复数z1=1+i,z2=3-i,则z1·z2=( )A.4+2 i B. 2+ i C. 2+2 i D.3
13.设z1,z2,z3三点适合条件 _(1)+(z)_(2)+(z)_(3)=0 |(z)_(1)|=|(z)_(2)|=|(z)_(3)|=1, 证明z1,
[题目]如果复数z1,z2,z3满足等式 dfrac (({z)_(2)-(z)_(1))}(({z)_(3)-(z)_(1))}=dfrac (({z)_(1
1.设 ^2+(y)^2+(z)^2-z=0, 求 dfrac ({a)^2z}(a{y)^2}
如果^2+(z)^2=ln dfrac (z)(x),则^2+(z)^2=ln dfrac (z)(x)( ).A.^2+(z)^2=ln dfrac (
5.设z_(1)及z_(2)是两复数.求证:(1)|z_(1)-z_(2)|^2=|z_(1)|^2+|z_(2)|^2-2mathrm(Re)(z_(1)ov
椭球面^2+2(y)^2+(z)^2=1上平行于平面^2+2(y)^2+(z)^2=1的切平面方程( )^2+2(y)^2+(z)^2=1^2+2
( A ) = (x,y,z)|{x)^2+(y)^2+(z)^2=(a)^2,zgeqslant 0} ( B ) = (x,y,z)|{x)^2+(y)^
曲线 ) (x)^2-(y)^2+(z)^2=0 z=1 .曲线在坐标面上的投影曲线方程为______A.B.C.D.
(z)=(z)^2+dfrac (1)({z)^2-1},则其解析区域为()(z)=(z)^2+dfrac (1)({z)^2-1}(z)=(z)^2+dfra